Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

分析：根据题意，我们可以要找出三个数相加等于0的这样的一个集合，所以采用二维数组存储。

首先抽取一个变量出来，该变量从左往右递归遍历，递归的同时设置两个变量，让其一个从第一个变量的右边，一个从数组的末端，同步的向中间遍历，有点类似于快速排序的判断方式，

依次递归，直到第一个变量条件终止为止。

Code(c++):

class Solution {
public:
    vector
   
     > threeSum(vector
    
      &nums) { vector
     
       > result; sort(nums.begin(), nums.end()); for(int i = 0; i < nums.size(); i++){ if(i > 0 && nums[i] == nums[i-1]) continue; threeNumber(nums, result, i); } return result; } //return vector
      
        > results void threeNumber(vector
       
         &nums, vector
        
          > &results, int curIndex) { int i = curIndex + 1; int j = nums.size()-1; while(i < j){ if(nums[curIndex] + nums[i] + nums[j] < 0) i++; else if(nums[curIndex] + nums[i] + nums[j] > 0) j--; else{ vector
         
           v; v.push_back(nums[curIndex]); v.push_back(nums[i]); v.push_back(nums[j]); results.push_back(v); i++; j--; while(i < j && nums[i]==nums[i - 1]) i++; while(j > i && nums[j] == nums[j + 1]) j--; } } } };